$4x-x^2y+y^3=10$ Find the value of $\dfrac{dy}{dx}$ at the point $(1,2)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $\dfrac{2}{3}$ (Choice C) C $-4$ (Choice D) D $0$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $4x-x^2y+y^3=10$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} 4x-x^2y+y^3&=10 \\\\ \dfrac{d}{dx}(4x-x^2y+y^3)&=\dfrac{d}{dx}(10) \\\\ \dfrac{d}{dx}(4x)-\dfrac{d}{dx}(x^2y)+\dfrac{d}{dx}(y^3)&=0 \\\\ 4-\left(2x\cdot y+x^2\cdot\dfrac{dy}{dx}\right)+3y^2\cdot\dfrac{dy}{dx}&=0 \\\\ 4-2xy-x^2\cdot\dfrac{dy}{dx}+3y^2\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 4-2xy-x^2\cdot\dfrac{dy}{dx}+3y^2\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(3y^2-x^2)&=2xy-4 \\\\ \dfrac{dy}{dx}&=\dfrac{2xy-4}{3y^2-x^2} \end{aligned}$ Now we can plug the point $(1,2)$ into the expression for $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{2xy-4}{3y^2-x^2} \\\\ &=\dfrac{2(1)(2)-4}{3(2)^2-(1)^2} \gray{x=1,\,\,y=2} \\\\ &=\dfrac{0}{11} \\\\ &=0 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at the point $(1,2)$ is $0$.